256
Biomedical Signal and Image Processing
Now, assume that at angle θ, we send a number of parallel beams with different t’s
(as in Figure 13.8) and detect the amount of attenuation (hereafter “projection”) at the
other side of the tissue. Since for each t we obtain a projection, the resulting projection values can be thought of as a function of t, i.e., the projection function obtained
at angle θ can be represented as P (t). One might argue that since the amount of shift
θ
t is often discrete, the function P (t) should also be a discrete function. While this
θ
argument is true and, in reality, both t and P (t) are discrete, at this point, we restrict
θ
ourselves to continuous variables and functions and get back to this issue later on.
A typical P (t) is shown in Figure 13.8, where more attenuation is obtained in the
θ
middle. This is due to the fact that the beam has to travel more through the tissue in
middle points and therefore gets attenuated more.
Next, we try to see if we can estimate the tissue structure from the projection
function, i.e., estimate f(x, y) using P (t). Let us first explore the integral relation
θ
between these two functions again:
P t
q ( ) =
f (x , y) ds
(13.4)
( ,
q t
∫
) line
As can be seen in Equation 13.4, in order to estimate f(x, y) from P (t), one needs to
θ
solve an integral equation. Before doing so, we first explore special cases of Equation
13.4 that give us more insight to the nature of this equation. These special cases are
when θ = 0 (beams parallel with x-axis) and θ = π/2 (beams parallel with y-axis).
For θ = 0, we have ds = dy, and, therefore,
∫
+∞
P q =0 ( )
t =
f (x, y) ds =
∫
f (x , y) dy = P q = 0 ( x )
(13.5)
(q =0 j ) lines
−∞
Equation 13.5 is a single variable integral over y, and the resulting projection function
has only one variable x. For θ = π/2, ds = dx, and, as a result,
+∞
P q p
= /2 ( )
t =
∫
f (x, y) ds =
∫
f (x , y) dx = P q p
= /2 ( y)
(13.6)
(q = p j ) lines
−∞
2
Equation 13.6 is also a single variable integral but this time over x and the resulting
projection function has only one variable y. These two special cases will help in our
attempt to solve the equations for f(x, y).
The main method to solve the preceding inverse problems applies the Fourier
transform (FT). This rather simple strategy has two steps: (1) Use P (t) to find or estimate
θ
F(u, v), i.e., FT of f(x, y) in (u, v) domain and (2) calculate inverse Fourier transform
(IFT) of F(u, v) (or its estimation) to obtain f(x, y). As can be seen, the second step is a
simple standard routine, and, as a result, conducting the first step, i.e., finding F(u, v)
using measurement of P (t), is the main part of tomography.
θ
Biomedical Signal and Image Processing
Now, assume that at angle θ, we send a number of parallel beams with different t’s
(as in Figure 13.8) and detect the amount of attenuation (hereafter “projection”) at the
other side of the tissue. Since for each t we obtain a projection, the resulting projection values can be thought of as a function of t, i.e., the projection function obtained
at angle θ can be represented as P (t). One might argue that since the amount of shift
θ
t is often discrete, the function P (t) should also be a discrete function. While this
θ
argument is true and, in reality, both t and P (t) are discrete, at this point, we restrict
θ
ourselves to continuous variables and functions and get back to this issue later on.
A typical P (t) is shown in Figure 13.8, where more attenuation is obtained in the
θ
middle. This is due to the fact that the beam has to travel more through the tissue in
middle points and therefore gets attenuated more.
Next, we try to see if we can estimate the tissue structure from the projection
function, i.e., estimate f(x, y) using P (t). Let us first explore the integral relation
θ
between these two functions again:
P t
q ( ) =
f (x , y) ds
(13.4)
( ,
q t
∫
) line
As can be seen in Equation 13.4, in order to estimate f(x, y) from P (t), one needs to
θ
solve an integral equation. Before doing so, we first explore special cases of Equation
13.4 that give us more insight to the nature of this equation. These special cases are
when θ = 0 (beams parallel with x-axis) and θ = π/2 (beams parallel with y-axis).
For θ = 0, we have ds = dy, and, therefore,
∫
+∞
P q =0 ( )
t =
f (x, y) ds =
∫
f (x , y) dy = P q = 0 ( x )
(13.5)
(q =0 j ) lines
−∞
Equation 13.5 is a single variable integral over y, and the resulting projection function
has only one variable x. For θ = π/2, ds = dx, and, as a result,
+∞
P q p
= /2 ( )
t =
∫
f (x, y) ds =
∫
f (x , y) dx = P q p
= /2 ( y)
(13.6)
(q = p j ) lines
−∞
2
Equation 13.6 is also a single variable integral but this time over x and the resulting
projection function has only one variable y. These two special cases will help in our
attempt to solve the equations for f(x, y).
The main method to solve the preceding inverse problems applies the Fourier
transform (FT). This rather simple strategy has two steps: (1) Use P (t) to find or estimate
θ
F(u, v), i.e., FT of f(x, y) in (u, v) domain and (2) calculate inverse Fourier transform
(IFT) of F(u, v) (or its estimation) to obtain f(x, y). As can be seen, the second step is a
simple standard routine, and, as a result, conducting the first step, i.e., finding F(u, v)
using measurement of P (t), is the main part of tomography.
θ